问题及解答

\[
\int_1^2 \frac{1}{x(1+x^n)}dx
\]

\[
\int_1^2 \frac{1}{x(1+x^n)}dx =\int_1^2 \frac{x^{n-1}}{x^n(1+x^n)}dx=\int_1^2 \frac{\frac{1}{n}dx^n}{x^n(1+x^n)},
\]

令 $t=x^n$, 则等于

\[
\begin{split}
\frac{1}{n}\int_{1}^{2^n}\frac{dt}{t(1+t)}&=\frac{1}{n}\int_{1}^{2^n}\Bigl(\frac{1}{t}-\frac{1}{1+t}\Bigr)dt\\
&=\frac{1}{n}\biggl(\ln t\biggl|_{1}^{2^n}-\ln(1+t)\biggl|_{1}^{2^n}\biggr)\\
&=\frac{1}{n}\Bigl[(n-1)\ln 2-\ln(1+2^n)\Bigr].
\end{split}
\]